How do you integrate #int 1/sqrt(-e^(2x) +9)dx# using trigonometric substitution?

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Answer 1 · 2016-03-14T02:26:04

#int(1/sqrt(-e^(2x)+9))=(ln(sqrt(9-e^(2x))-3)-x)/3#

Here is the corresponding right triangle from which you can extract all the parts for substitution:

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You need to find the following and then systematically replace all x terms (rectangular) with #theta# terms (trig):

x

#dx/(d theta)#

#sqrt(9-e^(2x))#

I'll start you off in the right direction:

Since

#sin(theta) =e^x/3 #

We have then that:

#x = ln(3*sin(theta))#

Now differentiate with respect to #theta# and split up the differentials.

You can then rewrite that entire integrand in terms of #theta# and solve using trigonometric substitution.

I'll give you one more piece and then I gotta sleep!

What is #cos(theta)?#

Well, referencing the triangle you can see that:

#cos (theta) = sqrt(9-e^(2x))/3#

This says that:

#sqrt(9-e^(2x)) =3*cos(theta)#

You'll still need to find #d theta#

So now you can systematically replace all rectangular terms with
trigonometric terms.

That's your job!